Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance case study question with answers
One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
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Answer: (a) 10 nC
V = Q/(4πε₀r), so Q = Vr × 4πε₀ = 900 × 0.10/(9 × 10⁹) = 1.0 × 10⁻⁸ C = 10 nC.
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Answer: (d) 15 cm
r = Q/(4πε₀V) = 9 × 10⁹ × 1.0 × 10⁻⁸/600 = 0.15 m. Since V ∝ 1/r, the 600 V surface is not midway between the 900 V and 300 V surfaces.
Answer and marking scheme
W = q(Vfinal − Vinitial) = 2.0 × 10⁻⁹ × (900 − 300) = 1.2 × 10⁻⁶ J. It does not depend on the path: the electrostatic force is conservative, so the work depends only on the end points.
- W = qΔV = 1.2 × 10⁻⁶ J (1)
- independent of path, since the force is conservative (1)
Answer and marking scheme
E = Q/(4πε₀r²) = 9 × 10⁹ × 1.0 × 10⁻⁸/(0.20)² = 2250 N C⁻¹ (equally, V/r = 450/0.20). It points radially outward, normal to the equipotential surface, towards lower potential.
- E = 2250 N C⁻¹ (2.25 × 10³ N C⁻¹) (1)
- radially outward, normal to the surface (1)
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